# Image, Range, and Matrix of a Linear Transformation

#### SiddharthThakur

##### New member
Question
Consider the linear transformation T(x1,x2,x3)= (2*x1 -2*x2- 4*x3 ,x1+2*x2+x3)
(a) Find the image of (3, -2, 2) under T.
(b) Does the vector (5, 3) belong to the range of T?
(c) Determine the matrix of the transformation.

Please see the attachement.it will be more clear to you.

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Staff member

#### Jameson

Staff member
Question
Consider the linear transformation T(x1,x2,x3)= (2*x1 -2*x2- 4*x3 ,x1+2*x2+x3)
(a) Find the image of (3, -2, 2) under T.
(b) Does the vector (5, 3) belong to the range of T?
(c) Determine the matrix of the transformation.

Please see the attachement.it will be more clear to you.
(a) The image of $(3, -2, 2)$ here is simply plugging in $x_1=3$, $x_2=-2$ and $x_3=2$ into the expression you are given.

(b) You want to solve $$\displaystyle A \hspace{1 mm} \vec{x} = \left[ \begin{array}{c}5 \\ 3 \end{array}\right]$$. You can do that by the following method.

$$\displaystyle \left[ \begin{array}{ccc} 2 & -2 & 4\\ 1 & 2 & 3 \end{array}\right] \left[ \begin{array}{c}x_1 \\ x_2 \\ x_3 \end{array}\right] = \left[ \begin{array}{c}5 \\ 3 \end{array}\right]$$

Change that to an augmented matrix and solve.

#### Jameson

Staff member
Question
Consider the linear transformation T(x1,x2,x3)= (2*x1 -2*x2- 4*x3 ,x1+2*x2+x3)
(a) Find the image of (3, -2, 2) under T.
(b) Does the vector (5, 3) belong to the range of T?
(c) Determine the matrix of the transformation.
Since the deadline for the assignment in question has passed, I will make some more comments on solving the problem so others can use the information in the future.

(а) $Т(3,-2,2)=(2(3)-2(-2)-4(2), 3+2(-2)+2)=(2,1)$

(b) To answer this we can start by answering part (c). The transformation matrix of $T$ is $\left[ \begin{array}{ccc} 2 & -2 & 4\\ 1 & 2 & 3 \end{array}\right]$ To find if the vector $[5,3]$ is in the range of $T$ we can solve the following matrix equation:

$$\displaystyle \left[ \begin{array}{ccc} 2 & -2 & -4\\ 1 & 2 & 3 \end{array}\right] \left[ \begin{array}{c}x_1 \\ x_2 \\ x_3 \end{array}\right] = \left[ \begin{array}{c}5 \\ 3 \end{array}\right]$$

I will leave the computation to the reader, but if you row-reduce the augmented matrix from the above equation then you'll find that the the system is consistent, thus there exists a solution and the vector $[5,3]$ is in the range of $T$.

(d) The transformation matrix is composed of 3 column vectors in $\mathbb{R}^2$ and after noticing that there are two of them which are linearly independent it follows that $T$ is in fact onto $\mathbb{R}^2$. This can also be shown by using the fact that there is a pivot position in every row.
(e) We have three column vectors in $\mathbb{R}^2$, so at least one of them must be a linear combination of the other two so $T$ isn't one-to-one. We can see this fact when row-reducing the augmented matrix for part (b). The solution contains a free variable which means that there is more than one solution, which doesn't fit the definition of being one-to-one. Two ways to check this in general is the columns of the transformation matrix must be linearly independent and when row-reducing the augmented matrix to solve for any $\vec{b}$, there can't be any free variables.